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# Magnetism Study Guide (page 2)

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Updated on Sep 27, 2011

#### Example

In a particle accelerator, an electron is accelerated in a field of 5,000 gauss. If the speed of the electron is 3 · 106 m/s, find the magnetic force on the electron if the electron enters the field at an angle of 30° with respect to the magnetic field.

#### Solution

Set the quantities we are given and the unknown in the equation of the magnetic field.

B = 5,000 G = 5,000 G · 1 T/104 G = 0.5 T

v = 3 · 106 m/s

α = 30°

F = ?

F = q · v · B · sin α

F = –1.6 · 10–19 C · 3 · 106 m/s · 0.5 T · sin 30°

F = –120 · 10–15 N

F = –1.2 · 10–13 N

## The Magnetic Force on an Electrical Current in a Magnetic Field

As we have seen, a charge is acted upon by a magnetic field. But a charge in motion creates an electrical current, so some interaction must take place between a wire carrying current and an applied magnetic field.

We will consider a long wire of length L carrying a current I. If the free electrons are conducting electricity through the wire then once we have a magnetic field, each electron will sense the field and interact with it.

In a time Δ t, N electrons will be passing through the cross-sectional area with a charge of N · qe. Let's consider a wire with a circular cross-sectional area and electrons moving with an average velocity v (this is also called drift velocity). See Figure 15.4.

Then, by the definition of the electrical current:

I = N · qe/ Δt

where N is the density of charges moving through the conductor.

Solving for the total charge:

I · Δt = N · qe

Also, the volume that the N charges occupy:

v = A · L = A · (v · Δt)

And then the time can be found to be:

Δt = V/A · v

Introducing this is the current formula:

I · V/ A · v = N · qe

We also know that for a charge, the magnetic force is:

F = q · v · B · sin α

Then for N charges, the magnetic force will be calculated the same, but the charge is N · qe. Replacing the charge and simplifying, we get:

F = I · L · B · sin α

In this formula, the angle α is the angle between the electrical current and the applied magnetic field.

This result is very similar to the magnetic force on a moving charge that we worked with at the beginning of this lesson, although the place of the q · v product is taken by the I · L. The same rules, the right-hand rule or the corkscrew rule, can be employed to find the direction of the resultant force, and you can probably already draw the conclusion that the force will be perpendicular to the plane formed by the current and magnetic field directions, similar to the conclusion of the first part. However, determination of the magnetic force direction is made easier by the fact that we work directly with current and not charge: The right-hand thumb will take the direction of the electric current, the fingers point in the direction of the magnetic field, and the magnetic force will exit the palm perpendicularly.

#### Example

In Figure 15.5, the same wire is conducting the same current, but the direction of the field and of the magnetic field are different in the two images. Determine the direction of the magnetic force.

#### Solution

As shown by the figure below, the right hand rule is applied for each of the drawings:

1. The current is in the direction of the y-axis, magnetic field is in the direction of the z-axis. The resultant magnetic force vector is in the direction of the direction of the x-axis.
2. Both the current and the magnetic field are in the direction of the z-axis, and so the angle between them is α = 0°. The product is zero. So, no magnetic force is acting on the conductor in Figure 15.6.
3. In Figure 15.7, the current is in the –x-axis direction, and the magnetic field is on z-axis direction. If you use the right-hand rule or the corkscrew rule, you determine a y direction for the magnetic force.

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